Modul IV
Separable Variable Diff.Eq.
(continued)
Contoh 4-1.
Hitunglah : x sin
Contoh 4-2
Hitunglah:
xe y dx  ydy  0
dy
x4

dx
x3
dengan initial condition y(1)=2
Contoh 4-3
Hitunglah :
dy
x

dx
y
subjec pd initial condition y(4)=3
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Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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Dgn initial condition y(4)=3,maka constanta c dpt dihitung yaitu substitusikan nilai x=4
dan y=3 kedalam pers.
terachir,thus
Dan lingkaran yg melalui titik (4,3) terlihat pada Gamb.3.3.
Soal-soal latihan 4:
4-1Hitunglah :
axy'by  xyy'
4-2.Hitunglah :
y
4-3 Hitunglah :
2(1  y ) xydx  1  x 2 1  y 2 dy  0
4-4Hitunglah :
2

 2 y  5 dx  dy  0
2



sec .2  tan d  sec .2  tan d  0
Jawab:
Disini penulis coba memecahkan soal2 diatas dengan mengacu pada ketentuan
Definisi 2 diatas:
Jawab:4-1
4-1. axy   by  xyy 
Kita atur dahulu untuk memudahkan perhitungan ,yaitu:
axy   by  xyy 
xyy   axy   by
( xy  ax) y   by
Rubah kebentuk ”pers.diff.separable variable” dengan
dy
 y  ,maka:
dx
dy
 by
dx
( y  a) b
 dx
y
x
( xy  ax )
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Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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Dengan metoda Integral didapat:
( y  a)

y
a
 dy   y dy 

b
dx
x
1
dx
b
y  c1  a log y  c 2  b log x  c 3
y  a log y  b log x  c
y  a log y  b log x  c
y  log y a  log x b  c
 y  log y a x b  c.....( jawab)
Jawab 4-2:
4-2. ( y  2 y  5)dx  dy  0
Rubah kebentuk ”pers.diff.separable variable”,yaitu:
2
( y 2  2 y  5)dx  dy  0
1
dy  dx....(i )
( y 2  2 y  5)
Dengan metoda Integral didapat:


1
dy   dx
( y 2  2 y  5)
1
dy   dx.....(ii )
( y  1) 2  2 2
1
du  dx
u2  a2
Kita tahu dari daftar integral bahwa: 1
u
tan 1  c 2  x  c1
2
a
Maka:
1 1 y  1
tan (
)  c 2  x  c1
2
2
:
1
y 1
x  c  tan 1 (
)
2
2
Setelah diatur kembali didapat pers.diff yang dicari,yaitu:
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2 x  c  tan 1 (
y 1
)
2
y 1
 tan( 2 x  c)
2
 y  2 tan( 2 x  c)  1....( jawab)
3-3 Pers.Diff. Separable Variable dengan Initial Value Problem
Perhatikan contoh-contoh soal dibawah ini:
4-1. 2(1  y ) xydx  (1  x )(1  y )dy  0
Rubah kebentuk ”pers.diff.separable variable”,yaitu:
2
2
2
2(1  y 2 ) xydx  (1  x 2 )(1  y 2 )dy  0
.
(1  y 2 )
2x
dy

dx  0
(1  y 2 ) y
(1  x 2 )
Item pertama kita rubah menjadi:
(1  y 2 )
2x
dy 
dx  0
2
(1  y ) y
(1  x 2 )
(1  y 2 )  2 y 2
2x
dy

dx  0
(1  y 2 ) y
(1  x 2 )
(1  y 2 )  2 y 2
2x
dy 
dx  0
2
(1  y ) y
(1  x 2 )
 1  2 y 2 )dy  2 x 2 dx  0
y (1  y )
(1  x )
log y  log( 1  y 2 )  log( 1  x 2 )  c
log y (1  x 2 )  log c  log( y 2  1)
 y (1  x 2 )  c( y 2  1).....( jawab)
4-2. sec  tan d  sec  tan d  0...(i )
Rubah kebentuk ”pers.diff.separable variable”,yaitu:
2
2
sec 2  tan d  sec 2  tan d  0...(i )
. sec 2 
tan 
d 
sec 2 
d  0
tan 
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Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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Kita tahu dari daftar integral bahwa: d (tan f )  (sec 2 f )d ( f ) ,maka:
log(tan  )  log(tan  )  0
tan  tan   c....( jawab)
4-1 Separable equations
We solved the first order linear homogeneous equation
….(4-1)
by dividing both sides of the equation by y(t) to obtain the equivalent equation
….(4-2)
and observing that Equation (2) can be written in the form
……..(4-3)
We then found ln|y(t)|, and consequently y(t), by integrating both sides of (4-3). In an
exactly analogous manner, we can solve the more general differential equation
………(4-4)
where f and g are continuous functions of y and t . This equation, and any other
equation which can be put into this form, is said to be separable. To solve (4-4), we first
multiply both sides by f (y) to obtain the equivalent equation
…..(4-5)
Then, we observe that (4-5) can be written in the form
……..(4-6)
where F(y) is any anti-derivative of f(y); i.e., F(y) =

f (y)dy. Consequently,
……….(4-7)
Where c is an arbitrary constant of integration ,and we solve for y=y(t) from (4-7) to find
the general solution of (4-4).
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Solution: Multiplying both sides of this equation by y2
gives
In addition to the differential equation (4), we will often impose an initial condition on y(t)
of the form y(tn) = Yn .The differential equation (4-4) together with the initial condition y(t)
=
Yn is called an initial value problem. We can solve an initial value problem two
different ways. Either we use the initial condition y(tQ) = yn to solve for the constant c
in (4-7), or else we integrate both sides of (4-6) between tQ and t to obtain that
……(4-8)
If we now observe that
y
F(y) - F(yQ) =

f(r)dr………. (4-9)
y0
then we can rewrite (4-8) in the simpler form
y

y0
t
f(r)dr =

g(s)ds. ..(4-10)
t0
Example 4-3: Find the solution y{t) of the initial value
Problem
ey
dy
- (t+t3) = 0, yd) = 1.
dt
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Solution:
Method (i).
From Example 2, we know that the general solution of this equation is y = In(t2 /2 + t4 /4
+ c).
Setting t = 1 and y = 1 gives 1 = In(3/4 + c), or
c = e - 3/4. Hence, y(t) = ln(e-3/4 + t2/2 + t4/4).
Method (ii). From (4-10),
y

t
e dr =
1

(s+s3)ds.
1
Consequently,
Example 4-4
Solve the initial value problem
dy
= 1 + y2 ,y(0) = 0.
dt
Solution:
Divide both sides of the differential equation by
1 + y2 to obtain the equivalent equation :
1
dy
1
2
(1  y ) dt
Then, from (4-10
Consequently, arc tan y = t, and y = tan t.
The solution y = tan t of the above problem has the disturbing property that it goes to
±  at t = ±π/2. And what's even more disturbing is the fact that there is nothing at all in
this initial value problem which even hints to us that there is any trouble at t = ±π/2. The
sad fact of life is that solutions of perfectly nice differential equations can go to infinity in
finite time. Thus, solutions will usually exist only on a finite open interval a < t < b, rather
than for all time. Moreover, as the following example shows, different solutions of the
same differential equation usually go to infinity at different times.
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From the equation (4-10)
Consequently, arc tan y -arc tan 1 = t, and y = tan( t+π/4).This solution exists on the
open interval -3π/4<t<π/4.
Example 6;
y
Find the solution y(t) of the initial value problem
dy
+ (1 + y2)sin t = 0,
dt
y{0) = 1.
Solution:
Dividing both sides of the differential equation by (1 + y2) gives:
1/2 ln(l+y2) – 1/2 In 2 = cos t - 1.
Solving this equation for y(t) gives
To determine whether we take the plus or minus branch of the square root, we note that
y(0) is positive. Hence,
This solution is only defined when
Since the logarithm function is monotonic increasing, we may take logarithms of both
sides of (4-11) and still preserve the inequality. Thus, 4 sin t/2 < In 2, which implies that
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Therefore y(t) only exists on the open interval (-a,a) where a = 2 arc sin[
In 2/2].
Now, this appears to be a new difficulty associated with nonlinear equations, since y(t)
just "disappears" at t = ±a, without going to infinity.
However, this apparent difficulty can be explained quite easily, and moreover, can even
be anticipated, if we rewrite the differential equation above in the standard form
Notice that this differential equation is not defined when y = 0. Therefore, if a solution y{t)
achieves the value zero at some time t = t*, then we cannot expect it to be defined for t >
t*. This is exactly what happened here, since y(±a) = 0.
Example 4-7:
Solve the initial value problem
dy
= (l+y)t,
dt
y(0) = -1.
Solution:
In this case, we cannot divide both sides of the differential equation by 1 + y, since y(0)
= -1. However, it is easily seen that y(t) = -1 is one solution of this initial value problem,
and in Section 1,we show that it is the only solution. More generally, consider the initial
dy
=f(y)g(t), y(tQ) = y0 ,where f(y0) = 0. Certainly, y{t) = y0 is one solution
dt
of this initial value problem, and in Section 1,we show that it is the only solution if  f/
 y exists and is continuous.
value problem
Example 4-8:
Solve the initial value problem (l+ey)
dy
= cos t, y(π/2) = 3.
dt
Solution; From (4-10) ,
so that y + ey = 2 + e3 + sin t. This equation cannot be solved explicitly for y as a
function of t. Indeed, most separable equations cannot be solved explicitly for y as a
function of t. Thus, when we say that
y + ey = 2 + e3 + sin t
is the solution of this initial value problem, we really mean that it is an implicit, rather
than a explicit solution.
This does not present us with any difficulties in applications, since we can always find
y (t) numerically with the aid of a digital computer.
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Example 4-9: Find all solutions of the differential equation
dy
t

dt
y
Solution:
Multiplying both sides of the differential equation by y gives y dy/dt = -t. Hence
y2 + t2 = c2. (4-12)
Now, the curves (12) are closed, and we cannot solve for y as a single-valued function
of t. The reason for this difficulty, of course, is that the differential equation is not
defined when y = 0. Nevertheless, the circles y2 + t2 = c2 are perfectly well defined,
even when y = 0. Thus, we will call the circles y2 + t2 = c2 solution curves of the differential equation
dy
t
  . More generally, we will say that any curve defined by (4-7)
dt
y
is a solution curve of (4-4).
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